Hi r-help-boun...@r-project.org napsal dne 21.07.2010 17:46:02:
> > Hi R-community, > I have the code as follows,i Fitted model as follows > lbeer<-log(beer_monthly) > t<-seq(1956,1995.2,length=length(beer_monthly)) #beer_monthly contains 400+ > entries not needed > t2=t^2 > beer_fit_parabola=lm(lbeer~t+t2) beer_fit_parabola=lm(lbeer~t+I(t^2)) you can use poly(t,2) but coefficients are not directly applicable in equation. > > Below is not working for me. > Please help me in preparing the new data set for the below prediction > > > > predict(beer_fit_parabola,newdata=data.frame(t=seq(1995,1998,length=20),t2=seq > (1995,1998,length=20)) predict(beer_fit_parabola,newdata=data.frame(t=seq(1995,1998,length=20)) shall be enough Regards Petr > > #it is listing all 400+ entries ,but not 20 ahead prediction. > > Thanks In advance for your help > -- > View this message in context: http://r.789695.n4.nabble.com/Help-me-with- > prediction-in-linear-model-tp2297313p2297313.html > Sent from the R help mailing list archive at Nabble.com. > > ______________________________________________ > R-help@r-project.org mailing list > https://stat.ethz.ch/mailman/listinfo/r-help > PLEASE do read the posting guide http://www.R-project.org/posting-guide.html > and provide commented, minimal, self-contained, reproducible code. ______________________________________________ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
